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A bilinear optimal control problem applied to a time
dependent Hartree-Fock equation coupled with classical
nuclear dynamics
Lucie Baudouin
To cite this version:
A bilinear optimal control problem applied to a
time dependent Hartree-Fock equation coupled
with classical nuclear dynamics
Lucie
Baudouin
November 2004
Abstract: We study a problem of bilinear optimal control for the electronic wave function of an Helium atom by an external time dependent electric field. The behavior of the atom is modeled by the Hartree-Fock equation, whose solu-tion is the wave funcsolu-tion of the electrons, coupled with the classical Newtonian dynamics, corresponding to the motion of the nucleus. We prove the existence of a bilinear optimal control in the case when the position of the nucleus is known and also prove the corresponding optimality condition. Then, we detail the proof of the existence of an optimal control for the coupled system and complete the study giving a formal optimality condition to define the electric control.
Keywords : Hartree-Fock equation, optimal control, optimality condition. AMS Classification : 49J20, 35Q55.
1
Introduction
We are interested in a bilinear optimal control problem applied to the mathe-matical model of the behavior of a simplified chemical system, in fact an Helium atom, controlled by an external electric field. We describe the chemical system in terms of ordinary and partial differential equations using very classical ap-proximations of quantum chemistry.
unbounded, and a nonlinearity of Hartree type in the right hand side. We want precisely to study the optimal control of the wave function of the electrons only, the control being performed by the electric potential.
We are in fact considering the following coupled system : i∂tu+ ∆u + 1 |x − a(t)|u+ V1u= (|u| 2⋆ 1 |x|)u, in R 3× (0, T ) u(0) = u0, in R3 m d 2a dt2 = Z R3−|u(x)| 2 ∇ 1 |x − a|dx− ∇V1(a), in (0, T ) a(0) = a0, da dt(0) = v0 (1)
where V1is the external electric potential depending on space and time variables,
which takes its values in R and satisfy the assumptions: (1 + |x|2)−1 2V 1 ∈ L∞((0, T ) × R3), (1 + |x|2)−1 2∂ tV1 ∈ L1(0, T ; L∞(R3)), (1 + |x|2)−1 2∇V 1 ∈ L1(0, T ; L∞(R3)) and ∇V1 ∈ L2 0, T ; Wloc1,∞(R3). (2)
We will define later on the optimal control problem related to this system and recall the precise results of existence and regularity of the solution we need in the sequel. One can already find in reference [2] the study of existence and regularity of solutions to this coupled system.
The Cauchy problem for this kind of non-adiabatic approximation of the general chemical Schr¨odinger equation has also been studied in the particular case when the atom is subjected to a uniform external time-dependent electric field I(t) such that in equation (1), one has V1 = −I(t) · x as in reference [5].
The authors remove the electric potential from the equation using a change of unknown function and variables (gauge transformation given in [8]). From then on, they have to deal with the nonlinear Schr¨odinger equation with only a time dependent coulombian potential.
We work in R3 and throughout this paper, we use the following notations:
∇v = ∂v ∂x1 , ∂v ∂x2 , ∂v ∂x3 , ∆v = 3 X i=1 ∂2v ∂x2i , ∂tv= ∂v ∂t,
Reand Im are the real and the imaginary parts of a complex number, h . , . i stands for the scalar product in an Hilbert space
W2,1(0, T ) = W2,1(0, T ; R3), for p ≥ 1, Lp= Lp(R3) and
We also define H1 = v∈ L2(R3), Z R3(1 + |x| 2 )|v(x)|2dx <+∞ H2 = v∈ L2(R3), Z R3(1 + |x| 2)2 |v(x)|2dx <+∞ .
One can notice that H1and H2are respectively the images of H1and H2under
the Fourier transform.
On a mathematical point of view, the optimal control problem consists in minimizing a cost functional depending on the solution of a state equation (here, a coupled system of partial differential equations) and to characterize the min-imum of the functional by an optimality condition. One will see in the sequel that even if we can prove existence of an optimal control for system (1), we cannot justify the optimality condition we formally obtain. However, the pro-cess will be described and fully proved in the following section in the simpler situation where the position of the nucleus is known at every moment.
Let (u, a) be a solution of system (1) where the external electric field V1 is
the control, and u1∈ L2 be a given target. We define the cost functionnal J by
J(V1, u) = 1 2 Z R3|u(T ) − u 1|2dx+ r 2kV1k 2 H
where r > 0 is a weight affecting the control cost and H =nV, 1 + |x|2−12
V ∈ H1(0, T ; W) and ∇V ∈ L2 0, T ; W1,∞o
where W is an Hilbert space which satisfies W ֒→ W1,∞(R3). The problem is:
Can one find a minimizer V1∈ H for inf{J(V, u), V ∈ H} ?
Remarks: 1) For example, in the space H, we can choose the Hilbert space W = H3⊕ Span{ψ1, ψ2, ..., ψm}
where m ∈ N, and for all i ∈ [[1, m]], ψi∈ W1,∞(R3) \ H3(R3).
2) In H, we can replace the hypothesis on ∇V by ∇V ∈ L20, T ; W1,∞ loc
as in assumption (2). Indeed, since we do not use any hypothesis on ∇V1 to prove
that the solution a is bounded in C([0, T ]) (see [2]), then we do not need any information on ∇V1at infinity in R3. We will give details later on.
We can actually prove the following theorem :
Theorem 1. There exists an optimal control V1∈ H such that
One can notice that we first need an existence result for a solution of the coupled system (1) in order to be able to formulate the bilinear optimal control problem . We have already proved one, in reference [2] (also in [1]), actually with a more general hypothesis on V1. Indeed, we have
Theorem 2. We assume that T is a positive arbitrary time and (1 + |x|2)−1V 1 ∈ L∞((0, T ) × R3), (1 + |x|2)−1∂ tV1 ∈ L1(0, T ; L∞(R3)), (1 + |x|2)−1∇V 1 ∈ L1(0, T ; L∞(R3)) and ∇V1 ∈ L2(0, T ; Wloc1,∞(R3)). (3)
If u0∈ H2∩ H2, a0, v0∈ R, then system (1) has at least a solution
(u, a) ∈ W1,∞(0, T ; L2) ∩ L∞(0, T ; H2∩ H2)
× W2,1(0, T ). Moreover, for any solution of (1) in this class, if ρ0>0 is such that
(1 + |x|2)−1 V1 W1,1(0,T,L∞)≤ ρ0,
then there exists R > 0 depending on ρ0such that kakC([0,T ])≤ R and if ρ1>0
is such that V1 1 + |x|2 W1,1(0,T,L∞ ) + 1 + |x|∇V12 L1(0,T,L∞ ) + k∇V1kL2(0,T ;W1,∞(B R))≤ ρ1 then there exists a non-negative constant K0
T,ρ1 depending on the time T , on ρ1, on ku0kH2
∩H2 and on |a0|, |v0|, such that: kukL∞(0,T ;H2∩H 2)+ k∂tukL∞(0,T ;L2)+ m da dt L∞(0,T ) + m d2a dt2 L1(0,T ) + sup t∈[0,T ] Z R3 |u(t, x)|2⋆ 1 |x| |u(t, x)|2 1 2 ≤ KT,ρ0 1. (4)
One can notice that if V1∈ H then it satisfies assumptions (2) and (3), and
we have at least a solution to equation (1) with Theorem 2. The optimal control problem is then well defined.
The reader may also notice that we do not give any uniqueness result for the coupled system (1) in Theorem 2. Indeed, even if we are convinced that the solution in this class is unique, we do not have a proof of this result up to now. Actually, E. Canc`esand C. Le Bris give a proof of existence and uniqueness of solutions for the analogous system without electric potential in [5]. Of course, the method for proving uniqueness used in this article cannot be applied here because the Marcinkiewicz spaces which are used do not suit the general electric potential V1 satisfying (3).
it is a main obstruction to the obtention of an optimality condition.
The next section presents the study of the situation where the position of the nucleus is known, instead of being the solution of an ordinary differential equation coupled to the Hartree-Fock equation. Without any coupling, the problem comes down to the difficulty of dealing with a nonlinear Schr¨odinger equation. In section 3, we give the proof of Theorem 1 and a formal optimality condition.
2
Non-Linear Schr¨
odinger Equation
Before studying the optimal control problem linked with the coupled situation described in the introduction, we will consider the position a(t) of the nucleus as known at any time t ∈ [0, T ] and forget the second equation in (1). Of course, this is too restrictive for the study of the control of chemical reactions by an external electric potential, but this section is only a first step in the study of the more realistic coupled situation. Moreover, in this present case, we will give a full result for the optimal control problem described further, from the existence of an optimal control to the proof of a necessary optimality condition.
2.1
Existence, uniqueness and regularity of solution
We consider the following non-linear Schr¨odinger equation : i∂tu+ ∆u + 1 |x − a|u+ V1u= (|u| 2⋆ 1 |x|)u, R 3× (0, T ) u(0) = u0, R3 (5)
where V1takes its values in R and we make the following assumptions:
a ∈ W2,1(0, T ) is known (1 + |x|2)−1V 1 ∈ L∞((0, T ) × R3), (1 + |x|2)−1∂ tV1 ∈ L1(0, T ; L∞) and (1 + |x|2)−1∇V 1 ∈ L1(0, T ; L∞(R3)). (6)
We have to underline that one can find in reference [7] the proof of existence, uniqueness and regularity for the analogous equation without the electric po-tential V1. This paper also deals with the more general case of an atom with
more than two electrons. We draw the reader’s attention on the fact that one of the main difficulty we encounter in the situation we are interested in is the coexistence of two potentials whose singularities are non-comparable.
to equation (5) and its proof is given in reference [2]. We first consider the linear Schr¨odinger equation
i∂tu+ ∆u + 1 |x − a|u+ V1u= 0, R 3× (0, T ) u(0) = u0, R3,
we set ρ > 0 and α > 0 such that V1 1 + |x|2 W1,1(0,T,L∞) + 1 + |x|∇V12 L1(0,T,L∞) ≤ ρ and d2a dt2 L1(0,T ) ≤ α (7)
and we have the following result:
Theorem 3. Let the initial data u0 belongs to H2 ∩ H2 and the electric
potential V1and the position a of the nucleus satisfy assumption (6). We define
the family of Hamiltonians {H(t), t ∈ [0, T ]} by H(t) = −∆ − 1
|x − a(t)|− V1(t). Then, there exists a unique family of evolution operators {U(t, s), s, t ∈ [0, T ]} (the so called propagator associated with H(t)) on H2∩ H
2 such that for u0∈
H2∩ H2 we have
(i) U(t, s)U (s, r)u0= U (t, r)u0 and U (t, t)u0= u0 for all s, t, r ∈ [0, T ];
(ii) (t, s) 7→ U(t, s)u0 is strongly continuous in L2 on [0, T ]2 and
U(t, s) is an isometry on L2,that is kU(t, s)u0kL2 = ku0k
L2; (iii) U (t, s) ∈ L(H2∩ H2) for all (s, t) ∈ [0, T ]2 and (t, s) 7→ U(t, s)u0
is weakly continuous from [0, T ]2 to H2∩ H2; moreover,
there exists MT,α,ρ>0 such that: ∀t, s ∈ [0, T ], ∀f ∈ H2∩ H2,
kU(t, s)fkH2∩H
2≤ MT,α,ρkfkH2∩H2. (iv) The equalities i∂tU(t, s)u0= H(t)U (t, s)u0
and i∂sU(t, s)u0= −U(t, s)H(s)u0 hold in L2.
Now, Theorem 3 is the main ingredient to prove the following result of existence along with a Picard fixed point theorem.
Theorem 4. Let T be a positive arbitrary time and α and ρ satisfy (7). Under assumption (6), and if we also assume u0∈ H2∩ H2, then equation (5)
has a unique solution
u∈ L∞(0, T ; H2∩ H2) with ∂tu∈ L∞(0, T ; L2)
and there exists a real constant C > 0 depending on T , u0, α and ρ such that:
kukL∞
(0,T ;H2
We draw the reader’s attention to the uniqueness of the solution of (5) in this result. Thus, we can correctly define an optimal control problem on equa-tion (5), the control being the external electric potential V1 and the solution u.
From now on, we may denote 1
|x − a| by V0 and we mean a ∈ W
2,1(0, T ).
Theorem 3 is also useful to give a meaning to the equations we will encounter in the sequel. More precisely, we consider the general equation
i∂tv+ ∆v + V0v+ V1v= f (v), R3× (0, T )
v(0) = v0, R3 (8)
and we give the following result:
Proposition 5. Let T be a positive arbitrary time. Under assumption (6), if we assume v0 ∈ L2 and if we also assume that there exists C, CM >0 such
that for all u, v ∈ C([0, T ]; L2),
kf(u)kL1(0,T,L2)≤ CT (kukC([0,T ];L2)+ 1) and when kukC([0,T ];L2)≤ M and kvkC([0,T ];L2)≤ M,
kf(u) − f(v)kL1(0,T,L2)≤ CMTku − vkC([0,T ];L2), then equation (8) has a unique solution v ∈ C([0, T ]; L2).
The proof uses a Picard fixed point theorem on the functional Φ defined on the space C([0, T ]; L2) by Φ : v 7→ v 0− i Z . 0 U(·, s)f(v(s)) ds where U is the propagator of Theorem 3.
2.2
Optimal control problem
On the evolution system (5), we define an optimal control problem which reads as follows: if u1∈ L2 is a given target, find a minimizer V1∈ H for
inf{J(V ), V ∈ H} (9)
where the cost functionnal J is defined by J(V1) = 1 2 Z R3|u(T ) − u 1|2dx+ r 2kV1k 2 H, r >0. (10) There, H =nV, 1 + |x|2− 1 2 V ∈ H1(0, T ; W )o
where W is an Hilbert space such that W ֒→ W1,∞(R3) and in (10), u is the
Remarks: 1) One can notice that if V1 belongs to H, then it satisfies (6) and
we can apply Theorem 4 that gives a unique solution u.
2) This space H has been chosen here as an Hilbert space in order to have a differentiable norm.
3) This optimal control problem is a so-called “bilinear optimal control prob-lem” and the mapping control → state (V17→ u) is strongly nonlinear.
Let us now formulate the result on the bilinear optimal control problem. Theorem 6. There exists an optimal control V1∈ H such that
J(V1) = inf{J(V ), V ∈ H}
and for all δV in H, V1 satisfies
rhV1, δViH = Im Z T 0 Z R3 δV(x, t)u(x, t)¯p(x, t) dxdt (11) where u is solution of (5) and p is solution of the following adjoint problem, set in R3× (0, T ). i∂tp+ ∆p + V0p+ V1p= (|u|2⋆ 1 |x|)p + 2i(Im(up) ⋆ 1 |x|)u p(T ) = u(T ) − u1. (12)
Remark: If we substitute the Hilbert space H by a reflexive space satisfying assumption (6) in the definition of J and in (9), the existence of an optimal control can also be proved. Nevertheless, the proof of an optimality condition needs an Hilbert space.
A result of existence for a bilinear optimal control problem, governed by a Schr¨odinger equation with the same Hartree non-linearity (|u|2⋆ 1
|x|)u, has
also been given by E. Canc`es, C. Le Bris and M. Pilot in [6]. The au-thors deal with an electric potential homogeneous in space V1= −I(t) · x with
I ∈ L2(0, T ), while we take into account here a more general electric poten-tial optimal control. For instance, in the definition of H, we can consider the Hilbert space W = H3⊕Span{ψ
1, ψ2, ..., ψm} with m ∈ N, and for all i ∈ [[1, m]],
ψi ∈ W1,∞(R3) \ H3(R3). Then W ֒→ W1,∞ and this example enables us to
deal both with the particular case of [6] where V1(x, t) = −I(t) · x but for
I ∈ H1(0, T ) and with general electric potentials (1 + |x|2)−1 2V
1(t) ∈ H2(R3)
which are non-homogeneous in space.
We can also specify the optimality condition in the particular case where W = H3(R3) ⊕ Span{ψ
1} by an optimality system. We choose for instance
condition (11), we can get an optimality system that reads: r(I − ∂2 t)(I − ∆)Y1= (I − ∆)−1 Im(up)p1 + |x|2 in R3× (0, T ) ∂t(Y1− ∆Y1)(T ) = ∂t(Y1− ∆Y1)(0) = 0 in R3 r E−d 2E dt2 = Im Z R3 upp1 + |x|2dx in (0, T ) dE dt (T ) = dE dt(0) = 0. where V1(x, t) = 1 + |x|2 1 2 (I − ∆)−1 Y
1(x, t) + E(t) and p is the solution
of the adjoint equation (12). The proof when W = H3 can be read for the
problem of optimal control for the linear Schr¨odinger equation in reference [3], the only changes being the adjoint equation solved by p and the absence of E.
The proof of Theorem 6 is divided in two steps. Existence of an optimal control can be treated first while the optimality condition requires the proof of the continuity and the differentiability of J. The regularity result of Theorem 4 is strongly needed for proving this differentiability result.
2.2.1 Existence of an optimal control
We will prove here the existence of an electric optimal control minimizing the cost functional. Indeed, we are going to prove:
∃ V1∈ H such that J(V1) = inf{J(V ), V ∈ H}.
Remark: The structure of the proof given in reference [3], for a bilinear optimal control problem defined on the linear Schr¨odinger equation, is analogous to the one we will follow here.
We consider a minimizing sequence (Vn
1 )n≥0 in H for the functional J:
inf H J(V ) = limn→∞J(V n 1). Since J(Vn 1) = 1 2 Z R3|u n(T ) − u1|2dx+ r 2kV n 1 k2H
where unis solution of (5) with potential V1= V1n, we then obtain that (V1n)n≥0
is bounded in H, independently of n. Up to a subsequence, we have Vn 1 ⇀ V1
weakly in H and
kV1kH ≤ limkV1nkH. (13)
The difficulty comes from the term kun(T ) − u1k2L2. More precisely, the point is to prove the weak convergence of un(T ) toward u(T ) in L2 and this is not
obvious. It will implie limkun(T ) − u1k2L2 ≥ ku(T ) − u1k2
for all t in [0, T ], un(t) −→ u(t) in L2, where u is the solution associated with V1, then lim n→+∞kun(T ) − u1k 2 L2 = ku(T ) − u1k2 L2, (14) and from (13) and (14) we obtain
J(V1) ≤ limJ(V1n) = inf V∈HJ(V ).
As V1 ∈ H, we get J(V1) = inf
H J and the existence of an optimal control is
proved.
We set F (u) = (|u|2⋆ 1
|x|)u and we consider wn = un− u solution of the
following equation:
i∂twn+ ∆wn+ V0wn+ V1nwn= F (un) − F (u) + u(V1− V1n), R3× (0, T )
wn(0) = 0, R3. (15)
We are going to prove that for all t in [0, T ], kwn(t)kL2 −→ 0.
In order to deal with the nonlinearity, we observe that we have, from Cauchy-Schwarz and Hardy’s inequality,
kF (u) − F (un)kL2 ≤ (|u|2⋆ 1 |x|)u − (|un| 2 ⋆ 1 |x|)un L2 ≤ (|u|2⋆ |x|1 )(u − un) L2 + (|u|2− |un|2) ⋆ 1 |x| un L2 ≤ 2kukL2k∇ukL2ku − unkL2 + 2kunkL2(k∇uk L2+ k∇unk L2) ku − unk L2 ≤ C(kuk2H1+ kunkH21)ku − unkL2. (16) Therefore, if we multiply equation (15) by wn integrate on R3 and take the
imaginary part, which means we calculate Im Z R3 (15).wn(x) dx, we obtain: d dt Z R3|wn| 2dx ≤ C kF (u) − F (un)kL2kwnkL2+ C Z R3|V n 1 − V1||u||wn| dx ≤ C kunk2H1+ kuk2H1 kwnk2L2 + C kVn 1 − V1kH Z R3|u|(1 + |x| 2)1 2|w n| dx
From Theorem 4, we have: kunkL∞(0,T ;H2∩H
2)+ k∂tunkL∞(0,T ;L2)≤ Cku0kH2∩H2 where C is independent of n since (Vn
1 )n≥0 is bounded in H. Then,
∀t ∈ [0, T ], kun(t)k2H1+ ku(t)k2
and we actually obtain (C denoting a generic constant depending on T ), d dt kwn(t)k 2 L2 ≤ C kwn(t)k2L2+ C Z R3|u(t)|(1 + |x| 2)1 2|w n(t)| dx (17)
Moreover, we will need the following compactness lemma (see reference [9] for its proof).
Lemma 7. Let X, B and Y be Banach spaces and p ∈ [1, ∞]. We assume that X ֒→ B ֒→ Y with compact embedding X ֒→ B.
If {fn, n ∈ N} is bounded in Lp(0, T ; X) and if {∂tfn, n ∈ N} is bounded
in Lp(0, T ; Y ) then {f
n, n ∈ N} is relatively compact in Lp(0, T ; B) (and in
C([0, T ]; B) if p = ∞).
Then, it has to be noticed that up to a subsequence we also have un −→ u
in C([0, T ]; H1
loc). Indeed, we can use Lemma 7 since (un)n≥0 is bounded in
L∞(H2∩ H
2) and (∂tun)n≥0 is bounded in L∞(L2). Then for all R > 0,
kwnkC([0,T ];L2(B(0,R))
n→∞
−→ 0 (18)
and on the other hand, for all t in [0, T ], Z B(0,R)c |wn(t)|2 1 + |x|2 dx !1 2 ≤ 1 1 + R2 1 2 kwn(t)kL2.
Thus, using Cauchy-Schwarz inequality, we can write Z R3|u(t)|(1 + |x| 2)1 2|w n(t)| dx ≤ Z B(0,R)|u(t)|(1 + |x| 2)1 2|w n(t)| dx + Z B(0,R)c|u(t)|(1 + |x| 2) |wn(t)| (1 + |x|2)1 2 dx ≤ ku(t)kH1kwn(t)kL2(B(0,R))+ 1 √ 1 + R2 ku(t)kH2kwn(t)kL2 ≤ C kwnkC([0,T ];L2(B(0,R))+ 1 √ 1 + R2 kwn(t)kL2 . (19)
We set En(t) = kwn(t)k2L2 + kwnkC([0,T ];L2(B(0,R)), where one can notice that kwnkC([0,T ];L2(B(0,R))does not depend on t. From (17) and (19), it satisfies
dEn dt (t) ≤ CEn(t) + C √ 1 + R2 p En(t).
then, from Gronwall lemma, we obtain that for all t in [0, T ], p En(t) ≤ CT eCT √ 1 + R2 + e CTkw nk 1 2 C([0,T ];L2(B(0,R)).
It means that since T is fixed and since we have (18), then for any ε > 0, there exists R > 0 and n0large enough in N such that
CT e CT √ 1 + R2 ≤ ε 2 and ∀n ≥ n0, e CTkw nk 1 2 C([0,T ];L2(B(0,R))≤ ε 2. We finally obtain that for all t in [0, T ], kwn(t)kL2
n→∞
−→ 0 and therefore, u is the solution of (5) in the sense of distributions and we have proved the existence of an optimal control V1 associated with the functionnal J. We then have to
write an optimality condition for V1.
2.2.2 Optimality condition
The usual way to obtain an optimality condition is to prove that the cost func-tional J is differentiable and to translate the necessary condition
DJ(V1)[δV1] = 0, ∀δV1∈ H
in terms of the adjoint state. Since J(V1) = 12ku(T ) − u1k2L2 + r2kV1k2H, as announced in the introduction, the main difficulty comes from the necessity to differentiate the state variable u with respect to the control V1, in order to
calculate the gradient DJ(V1). We postpone the proof of the following lemma.
Lemma 8. Let u be the solution of (5). The functional φ defined by φ: H → L2(R3)
V1 7→ u(T )
is differentiable and if z is the solution of the following equation, set in R3×
(0, T ): i∂tz+ ∆z + V0z+ V1z= −δV1u+ (|u|2⋆ 1 |x|)z + 2Re(uz ⋆ 1 |x|)u, z(t = 0) = 0 (20) we have z ∈ C([0, T ]; L2) and Dφ(V 1)[δV1] = z(T ).
We deduce from Lemma 8 that J is differentiable with respect to V1.
There-after, since Dφ(V1)[δV1] = z(T ), the condition
Remarks: 1) As for the study of the same bilinear optimal control problem for the linear Schr¨odinger equation one has read in reference [3], we can prove the differentiability of V17→ u(T ) with values in L2but we don’t know whether this
remains true if we consider the same mapping with values in H1 for example.
We think that the differentiability is not true anymore. Therefore, in the func-tional J, the first term cannot be replaced by a stronger norm of u(T ) − u1.
2) We can also underline the choice of H we made on purpose. As it is an Hilbert space, we can easily take the derivative of the norm k · kH in the
func-tional J.
Now, we consider the adjoint system (12): i∂tp+ ∆p + V0p+ V1p= (|u|2⋆ 1 |x|)p + 2i(Im(up) ⋆ 1 |x|)u in R 3× (0, T ) p(T ) = u(T ) − u1 in R3.
Using Proposition 5, one can prove that the equivalent integral equation has a unique solution p ∈ C([0, T ]; L2) since we have
(|u|2⋆ |x|1 )p + 2i(Im(up) ⋆ |x|1 )u L1(0,T ;L2) ≤ CT kpkC([0,T ];L2).
We then multiply equation (20) by p, integrate on [0, T ] × R3 and take the
imaginary part. We obtain: Im Z T 0 Z R3 (i∂tz+ ∆z + V0z+ V1z)p = Im Z T 0 Z R3−δV 1up +Im Z T 0 Z R3(|u| 2 ⋆ 1 |x|)zp + 2 Im Z T 0 Z R3 Re(uz ⋆ 1 |x|)up. Then z(0) = 0 implies Im Z T 0 Z R3 z i∂tp+ ∆p + (V0+ V1)p + Im i Z R3 z(T ) p(T ) = −Im Z T 0 Z R3 δV1up+ Im Z T 0 Z R3 z(|u|2⋆ 1 |x|)p +2 Z T 0 Z R3 (Im(up) ⋆ 1 |x|)Re(uz) and since p satisfies equation (12), we get
Using (22), the optimality condition (21) can be written : rhV1, δV1iH= Im Z T 0 Z R3 δV1up dxdt, ∀δV1∈ H.
The proof of Theorem 6 will be complete with the proof of Lemma 8. Proof of Lemma 8. Actually, we will first study the continuity of φ and then the differentiability. We recall the definition of the functional φ : if u is the solution of (5) with electric potential V1 in H, then
φ: H → L2(R3) V1 7→ u(T ).
According to Proposition 5 and to the properties of F , we consider the solution δu∈ C([0, T ]; L2) of the following equation set in R3× (0, T ):
i∂tδu+ ∆δu + V0δu+ (V1+ δV1)δu = −δV1u+ F (u + δu) − F (u)
δu(0) = 0. (23)
In order to prove the continuity of φ, we will prove that kδukL∞(0,T ;L2)= O(kδV1kH). Let us calculate Im
Z
R3
(23).δu(x) dx. Using the property (16) of F , we obtain d dt Z R3|δu| 2dx
≤ CkδV1kHkukH1kδukL2+ CkF (u + δu) − F (u)kL2kδukL2 ≤ CkδV1kHkukH1kδukL2+ C(kδu + uk
2 L2+ kδuk2 L2)kδuk2 L2 ≤ CkδV1kHkδukL2+ Ckδuk2 L2.
Indeed, the solution u of equation (5) and the solution u + δu of the same equation but with potential V1+ δV1, are bounded in L2. As δu(0) = 0 and
using Gronwall’s lemma, it follows
kδu(t)kL2≤ CT eCtkδV1kH , ∀t ∈ [0, T ].
Eventually, we get kδukC([0,T ];L2) = O(kδV1kH), the continuity of φ is proved and we will now prove the differentiability.
We first have to prove that z(T ) is well defined in L2 where z is solution of (20) and then, if we set w = δu − z, we will prove that
kw(T )kL2= o(kδV1kH)
Since we can prove the right hand side of equation (20) satisfies (|u|2⋆ 1 |x|)z + 2Re(uz ⋆ 1 |x|)u − δV1u L1(0,T ;L2) ≤ CT (kzkC([0,T ],L2)+ 1), then Proposition 5 gives a unique solution z ∈ C([0, T ]; L2) to equation (20).
Moreover, if we calculate Im Z
R3
(20).z(x) dx, we obtain from Hardy’s inequality: d dt Z R3|z| 2dx = −2Im Z R3 δV1uz dx+ 2 Z R3 (Re(uz) ⋆ 1 |x|)Im(uz) dx ≤ C kδV1kHkukH1kzkL2+ C Z R3 Z R3 |u(x)||z(x)| |x − y| |u(y)||z(y)| dxdy ≤ C kδV1kHkzkL2+ C k∇ukL2kzkL2 Z R3|u(x)||z(x)| dx ≤ C kδV1kHkzkL2+ C kzk2 L2. It implies kz(t)k2L2 ≤ CkδV1kH Z t 0 kz(s)k L2ds+ C Z t 0 kz(s)k 2 L2ds (24) and a Gronwall argument leads us easily to deduce that there exists a constants CT >0 such that
kz(t)kL2 ≤ CTkδV1kH, ∀t ∈ [0, T ]. (25) In order to simplify the right hand side of the equation solved by w = δu −z, we consider the source terms of equation (23) solved by δu:
F(u + δu) − F (u) − δV1u
= (|u + δu|2⋆ 1
|x|)(u + δu) − (|u|
2⋆ 1
|x|)u − δV1u = (|u|2⋆ 1
|x|)δu + 2Re(uδu ⋆ 1
|x|)(u + δu) + (|δu|
2⋆ 1
|x|)(u + δu) − δV1u and since z satisfies (20), we have finally the following right hand side
F(u + δu) − F (u) − δV1u− −δV1u+ (|u|2⋆ 1 |x|)z + 2Re(uz ⋆ 1 |x|)u = (|u|2⋆ 1 |x|)δu + 2Re(uδu ⋆ 1 |x|)(u + δu) + (|δu|2⋆ 1
|x|)(u + δu) − (|u|
Therefore, the equation satisfied by w in R3× (0, T ) is: i∂tw+ ∆w + V0w+ (V1+ δV1)w = −δV1z+ (|u|2⋆ 1 |x|)w + 2Re(uw ⋆ 1 |x|)u + 2Re(uδu ⋆ 1 |x|)δu + (|δu|2⋆ 1 |x|)(u + w + z) w(t = 0) = 0 (26)
Using Proposition 5, since the right hand side of equation (26) belongs to L1(0, T ; L2) and has the good properties, we can prove that there exists a unique
solution w ∈ C([0, T ]; L2). We can also formally calculate Im
Z
R3
(26).w(x) dx in the same way we did to prove (25). Since we have kδukL∞
(L2)= O(kδV1kH), we obtain: d dt(kwk 2 L2) ≤ C Z R3|δV 1||z||w| dx + C Z R3 (Re(uw) ⋆|x|1 )Im(uw) dx + C Z R3 Re(uδu ⋆ 1 |x|)Im(δu w) dx + C Z R3 (|δu|2⋆ 1 |x|)Im((u + z)w) dx ≤ CkδV1kHkzkH1kwkL2+ Ck∇ukL2kukL2kwk 2 L2 + C (k∇ukL2+ k∇zkL2) kδuk2 L2kwkL2 ≤ CkδV1kHkzkH1kwkL2+ Ckwk 2 L2+ CkδV1k2HkwkL2 + CkzkH1kδV1k2 HkwkL2 which means that for all t in [0, T ],
d
dt(kw(t)kL2) ≤ CkδV1kH(kz(t)kH1∩H1+ kδV1kH) + Ckw(t)kL2. (27) Since we want to prove that kwkL∞(0,T ;L2) ≤ CkδV1k2
H, we have to work
more on equation (20) in order to obtain an H1∩ H
1 estimate on z. Actually,
we could have directly proved with Theorem 3 and a Picard fixed point theorem that z ∈ C([0, T ]; H1∩ H
1). If we calculate Im
Z
R3(20).|x|
2z(x) dx, we obtain
Therefore, an integration on [0, t] and z(0) = 0 give k|x|z(t)k2L2 ≤ C Z t 0 k∇z(s)k 2 L2ds+ C Z t 0 kz(t)k 2 H1ds (28) + C Z t 0 kδV 1kHk|x|z(t)kL2ds
Now, as we need to estimate ∇z, we will calculate Re Z
R3
(20).∂tz(x) dx.
Before, we can notice that: Re Z R3(|u| 2⋆ 1 |x|)z∂tz= 1 2 d dt Z R3(|u| 2⋆ 1 |x|)|z| 2 − Z R3 (Re(u∂tu) ⋆ 1 |x|)|z| 2 and 2 Re Z R3 (Re(uz) ⋆ 1 |x|)u∂tz = d dt Z R3 (Re(uz) ⋆ 1 |x|)Re(uz) − 2 Z R3 (Re(uz) ⋆ 1 |x|)Re(z∂tu). After some calculations and integrations by parts, we obtain
d dt Z R3 V0|z|2+ Z R3 V1|z|2− Z R3|∇z| 2=Z R3 ∂tV0|z|2+ Z R3 ∂tV1|z|2 −2d dt Z R3 δV1Re(uz) + 2 Z R3 ∂t(δV1)Re(uz) + 2 Z R3 δV1Re(∂tu z) +d dt Z R3(|u| 2⋆ 1 |x|)|z| 2 − 2 Z R3 (Re(u∂tu) ⋆ 1 |x|)|z| 2 +2d dt Z R3 (Re(uz) ⋆ 1 |x|)Re(uz) − 4 Z R3 (Re(uz) ⋆ 1 |x|)Re(z∂tu). We recall here that V0(x, t) =
1
|x − a(t)| with a ∈ W
2,1(0, T ; R3) thus we
have |∂tV0(x, t)| = |∂t
a(t)|
|x − a(t)|2. We also remind Hardy’s inequality for u ∈
Therefore we obtain d dt Z R3|∇z(t)| ≤ d dt Z R3 (V0(t) + V1(t))|z(t)|2+ 2Re Z R3 δV1(t)u(t)z(t) + d dt 2 Z R3 (Re(u(t)z(t)) ⋆ 1 |x|)Re(u(t)z(t)) + d dt Z R3(|u(t)| 2 ⋆ 1 |x|)|z(t)| 2 + C k∇z(t)k2L1+ C kV1kHk|x|z(t)k2 L2 + C k∂tu(t)kL2k∇u(t)k L2kz(t)k L2 + C kδV1kH(ku(t)kL2+ k∂tu(t)kL2) k|x|z(t)kL2. We integrate this, between 0 and t ∈ [0, T ], using z(0) = 0,
u∈ L∞(0, T ; H2∩ H 2) and ∂tu∈ L∞(0, T ; L2). We obtain: k∇z(t)k2L2 ≤ Z R3 (V0(t) + V1(t))|z(t)|2+ 2 Z R3|δV 1(t)||u(t)||z(t)| + 2 Z R3(|u(t)||z(t)| ⋆ 1 |x|)|u(t)||z(t)| + Z R3(|u(t)| 2⋆ 1 |x|)|z(t)| 2 (29) + C Z t 0 k∇z(s)k 2 L2+ kz(s)k2H 1 ds+ C kδV1kH Z t 0 kz(s)k H1ds. We set E(t) = kz(t)k2H1+ kz(t)k2H 1= Z R3(1 + |x| 2 )|z(t, x)|2dx+ Z R3|∇z(t, x)| 2dx.
Moreover, we remind that we have (24) and (28) and adding this to (29), we get, for all t in [0, T ],
E(t) ≤ C kδV1kH Z t 0 p E(s) ds + C Z t 0 E(s) ds + Z R3 (V0(t) + V1(t))|z(t)|2+ C kδV1kHku(t)kH1kz(t)kL2 + C ku(t)kL2ku(t)k H1kz(t)k2 L2.
Then, we can prove that for all η > 0 there exists a constant Cη>0 such that
Z
R3
(V0(t) + V1(t))|z(t)|2≤ Cηkz(t)k2L2+ ηkz(t)k2
H1∩H
1. (30) Indeed, from Cauchy-Schwarz and Hardy’s inequalities, we have
an we obtain (30) from Young’s inequality. Consequently, if we choose η small enough, we obtain E(t) ≤ C kδV1kH Z t 0 p E(s) ds + C Z t 0 E(s) ds + Cηkz(t)k2L2+ C kδV1kHkz(t)kL2 and using (25), we get
E(t) ≤ C kδV1kH Z t 0 p E(s) ds + C Z t 0 E(s) ds + C kδV1k2H.
We recall that here again, C denotes various positive constants, depending only on the time T . We set
F(t) = kδV1kH Z t 0 p E(s) ds + Z t 0 E(s) ds + kδV1k2H. We have both E(t) ≤ C F (t) and dF dt(t) = E(t) + kδV1kH p E(t) ≤ C F (t) + C kδV1kH p F(t). Then, d dt e−CtpF(t)≤ e−CtCkδV
1kH and we obtain after an integration in
time:
∀t ∈ (0, T ), F (t) ≤ CkδV1k2H.
This implies that there exists a constant CT >0 such that
∀t ∈ (0, T ), E(t) = kz(t)k2 H1∩H
1≤ CTkδV1k
2 H.
Eventually, we have proved that, sup t∈[0,T ](k(1 + |x|)z(t)kL 2+ k∇z(t)k L2) kδV1kH→0 −→ 0.
Now, using this in (27), we obtain d
dt(kw(t)kL2) ≤ C kδV1k
2
H+ C kw(t)kL2
and applying Gronwall lemma we get: ∀t ∈ [0, T ], kw(t)kL2 ≤ CTkδV1k2
H.
Therefore, we have
kw(T )kL2= o(kδV1k
H)
3
Optimal control of the coupled system
We recall the coupled system (1) we are considering: i∂tu+ ∆u + u |x − a|+ V1u= (|u| 2⋆ 1 |x|)u, in R 3× (0, T ) u(0) = u0, in R3 m d 2a dt2 = Z R3−|u(x)| 2 ∇ 1 |x − a|dx− ∇V1(a), on (0, T ) a(0) = a0, da dt(0) = v0.
The electric potential V1takes its values in R and satisfy assumption (2):
(1 + |x|2)−1 2V 1 ∈ L∞((0, T ) × R3), (1 + |x|2)−1 2∂ tV1 ∈ L1(0, T ; L∞), (1 + |x|2)−1 2∇V 1 ∈ L1(0, T ; L∞) and ∇V1 ∈ L2 0, T ; Wloc1,∞. (31)
On this evolution system we define the following optimal control problem: if (u, a) is a solution of system (1) and if u1 ∈ L2 is a given target, find a
minimizer V1∈ H for
inf{J(V, u), V ∈ H} where the cost functionnal J is defined by
J(V1, u) = 1 2 Z R3|u(T, x) − u 1(x)|2dx+ r 2kV1k 2 H and H =nV, 1 + |x|2− 1 2 V ∈ H1(0, T ; W) and ∇V ∈ L2 0, T ; W1,∞o where W is an Hilbert space which satisfies W ֒→ W1,∞(R3).
We are now going to prove Theorem 1 and we will make, at the end of the proof, a remark about the obtaining an optimality condition.
Proof of Theorem 1. We consider a minimizing sequence (Vn
1)n≥0in H for
the functional J. It means that
inf{J(V, u), V ∈ H} = limn→∞J(Vn 1, un)
of i∂tun+ ∆un+ un |x − an| + Vn 1 un = (|un|2⋆ 1 |x|)un, in R 3× (0, T ) un(0) = u0, in R3 m d 2a n dt2 = Z R3−|u n(x)|2∇ 1 |x − an| dx− ∇Vn 1(an), on (0, T ) an(0) = a0, dan dt (0) = v0. (32) Since J(V1n, un) = 1 2 Z R3|un(T, x) − u 1(x)|2dx+ r 2kV n 1 k2H,
we then obtain that (Vn
1 )n≥0 is bounded in H, independently of n. Up to a
subsequence, we have Vn
1 ⇀ V1 weakly in H and
kV1kH≤ limkV1nkH.
The difficulty comes again from the term kun(T ) − u1k2L2. We will prove that the limit (u, a) of (un, an)n∈N is a solution of system (1) associated with V1.
If we consider a solution (un, an) of system (32), since the sequence of the electric
potentials (Vn
1 ) is bounded in H, we can apply Theorem 2 and obtain that the
sequence (un, an) is bounded in
W1,∞(0, T ; L2) ∩ L∞(0, T ; H2∩ H2)
× W2,1(0, T )
independently of n. We get an → a in L∞(0, T ) strongly and un⇀ u weak in
C([0, T ], L2).
Therefore, since the application u 7→ ku(T )−u1k2L2is lower semi-continuous, then un(T ) ⇀ u(T ) weak in L2(R3) implies
ku(T ) − u1k2L2 ≤ limkun(T ) − u1k2L2 and we finally obtain
J(V1, u) ≤ limJ(V1n, un) = inf
V∈HJ(V, u).
Since V1 ∈ H, that leads to J(V1, u) = inf
V∈HJ(V, u) and the existence of an
optimal control is proved.
Remark: As mentioned in the introduction, we can replace ∇V ∈ L2 0, T ; W1,∞
by ∇V ∈ L20, T ; Wloc1,∞
in the definition of H. Then, in the cost functional J, kV1kHhas to be replaced by (1 + |x|2)−1 2V 1 H1(0,T ;W)+k∇V1kL 2(0,T ;W1,∞(Bρ)) where Bρ = B(0, ρ) ⊂ R3 and the point is to choose ρ > 0 conveniently.
(1 + |x|2)−1 2V
1 ∈ H1(0, T ; W). Moreover, when we consider a minimizing
se-quence (Vn
1)n≥0 in H, as soon as J(V1n, un) is then bounded, for instance by
J(0, un), we obtain an a priori bound for k(1 + |x|2)−
1 2V
1kH1(0,T ;W) and then for kankC([0,T ]). Thus, if ρ is chosen large enough to satisfy kankC([0,T ])≤ ρ for
all n ∈ N, we will be able to proceed to the same proof as follows.
For clarity, we denote by (33) and (34) the two equations solved by u and a: i∂tu+ ∆u + u |x − a|+ V1u= (|u| 2⋆ 1 |x|)u, in R 3× (0, T ) (33) m d 2a dt2 = − Z R3|u(x)| 2 ∇ 1 |x − a|dx+ ∇V1(a), in (0, T ) (34) and we want to prove that the limit (u, a) of (un, an)n∈Nis a solution of (1).
Up to a subsequence, we have ∂tun → ∂tuand ∆un→ ∆u in D′((0, T )×R3).
Moreover, on the one hand, since V1is bounded in H, we have
Vn 1 (1 + |x|2)1 2 n≥0
bounded in H1(0, T ; W). Since the embedding W1,∞(R3) ֒→ L2
loc(R3) is
com-pact and since W ֒→ W1,∞(R3), then from Lemma 7, we get the local strong
convergence Vn 1 (1 + |x|2)1 2 n→+∞ −→ V1 (1 + |x|2)1 2 in L2(0, T ; L2loc).
On the other hand, (un)n≥0 is bounded in L∞(0, T ; H2) and since (un)n≥0 is
bounded in L∞(0, T ; H2)∩W1,∞(0, T ; L2), we have the local strong convergence
un n→+∞
−→ u in L∞(0, T ; L2loc).
Then, using Cauchy-Schwarz inequality, we can prove: Z T 0 Z BR V1n− V1 (1 + |x|2)1 2 un(1 + |x|2) 1 2 ≤ Z T 0 Z BR |Vn 1 − V1|2 (1 + |x|2) 1 2Z BR |un|2(1 + |x|2) 1 2 ≤ √T (Vn 1 − V1)(1 + |x|2)− 1 2 L2(0,T ;L2(B R)) kunkL∞(0,T ;H 2) ≤ CT (V1n− V1)(1 + |x|2)− 1 2 L2(0,T ;L2(B R)) n→+∞ −→ 0, and (1 + R2)1 2 Z T 0 Z BR V1 (1 + |x|2)1 2 (un− u) ≤ CT V1(1 + |x|2)− 1 2 L2 (0,T ;L2 )kun− ukL ∞ (0,T ;L2(B R)) ≤ CTkun− ukL∞(0,T ;L2(B R)) n→+∞ −→ 0 and for all ε > 0, there exists R > 0 such that
1 (1 + R2)1 2 Z T 0 Z Bc R |Vn 1un| + |V1u| (1 + |x|2)1 2 (1 + |x| 2) ≤ 2 √ T (1 + R2)1 2 V1n(1 + |x|2)− 1 2 L2(0,T ;L2)kunkL ∞ (0,T ;H2) ≤ CT (1 + R2)1 2 ≤ ε. Eventually, we obtain Vn 1un→ V1u in L1((0, T ) × R3).
Then, we have to work on the terms un |x − an| and |un|2⋆ 1 |x| un of
system (32). One can notice that (an)n≥0 is bounded in W2,1(0, T ). We then
have, up to a subsequence, the strong convergence an n→+∞
−→ a in L∞(0, T ).
We will check later that a, together with u, is a solution of coupled system (1). We set ϕ ∈ D((0, T ) × R3) which means in particular that Supp ϕ is a compact
set of (0, T ) × R3. We have un |x − an| = 1 |x − an|− 1 |x − a| un+ un |x − a|
and we will prove that in D′((0, T ) × R3), we have the following convergences
On the one hand, since Supp ϕ is compact, from Hardy’s inequality we have Z [0,T ]×R3 (un(t, x) − u(t, x))ϕ(t, x) |x − a(t)| dtdx ≤ Ckun− ukL∞(0,T ;H1(BR)) where Supp ϕ ⊂ (0, T )×BRand (un)n≥0being bounded in the space L∞(0, T ; H2)
∩W1,∞(0, T ; L2) gives the local strong convergence
un n→+∞ −→ u in C([0, T ]; Hloc1 ). Then kun− ukL∞(0,T ;H1(B R))→ 0 and we get un |x − a| n→+∞ −→ u |x − a| in D ′.
On the other hand, for the same reasons, we have Z [0,T ]×R3 1 |x − an(t)| − 1 |x − a(t)| un(t, x)ϕ(t, x) dtdx ≤ Z [0,T ]×R3 |un(t, x)||ϕ(t, x)||an(t) − a(t)| |x − an(t)||x − a(t)| dtdx ≤ kunkL∞(0,T ;H1(B R))|an− a|L∞(0,T ) Z Supp ϕ |ϕ(t, x)|2 |x − a(t)|2dtdx 1 2 ≤ C |an− a|L∞(0,T ) n→+∞ −→ 0 which means 1 |x − an|− 1 |x − a| un n→+∞ −→ 0 in D′.
which gives Z R3|ϕ| (|un|2⋆ 1 |x|)un− (|u| 2⋆ 1 |x|)u dx ≤ kun− ukL2(B R) Z R3|ϕ|(|u n|2⋆ 1 |x|) 2dx 1 2 (35) + kukL2 Z R3|ϕ| 2 ((|un| + |u|)|un− u| ⋆ 1 |x|) 2dx 1 2 . Next, from Hardy’s inequality, we have
(|un|2⋆ 1 |x|)(x) ≤ kunkL2k∇unkL2, ∀x ∈ R 3 and since ϕ ∈ D((0, T ) × R3), Z R3 ϕ(x)(|un|2⋆ 1 |x|) 2(x) dx 1 2 ≤ Ckunk2H1. (36) We will also need the following:
Lemma 9. Let r > 0, v ∈ H1 and v
n ∈ L2. If we assume that vn n→+∞ −→ 0 in L2 loc, then ∀|x| < r, Z R3 v(y)vn(y) |x − y| dy n→+∞ −→ 0.
Proof of Lemma 9. We set R > r and BR = {y ∈ R3,|y| < R}. From
Cauchy-Schwarz and Hardy’s inequalities, we obtain for all x such that |x| < r, Z R3 v(y)vn(y) |x − y| dy ≤ Z BR |v(y)||vn(y)| |x − y| dy+ Z Bc R |v(y)||vn(y)| |y − x| dy ≤ kvkH1kvnkL2(B R)+ 1 R− |x|kvkL2kvnkL2 ≤ C kvnkL2(B R)+ 1 R− r .
Moreover, if we set ε > 0, then there exists n0∈ N and R0>0 such that
C R0− r ≤ ε 2 and ∀n > n0, CkvnkL2(BR0)≤ ε 2. Thus, for all ε > 0 there exists n0∈ N such that for all n > n0,
Z R3 v(y)vn(y) |x − y| dy ≤ ε
We use this result to deal with the term Z R3 ϕ2((|un| + |u|)|un− u| ⋆ 1 |x|) 2.
Let t ∈ (0, T ) be fixed. Since Supp ϕ is compact, we apply Lebesgue’s theorem on a bounded domain to the sequence (fn(t))n∈N defined by
fn(x, t) =
(|un(t)| + |u(t)|)|un(t) − u(t)| ⋆
1 |x| 2 (x). Indeed, since un− u n→+∞ −→ 0 in C([0, T ]; L2 loc), u ∈ L∞(0, T ; H1) and un is
bounded in L∞(0, T ; H1) independently of n, using Lemma 9 we obtain that for
all t in [0, T ] and for all x in Supp ϕ, fn(x, t) → 0. Then, from usual estimates
we prove that |fn(t)| ≤ C ∈ L1loc(R3) and we finally get: ∀t ∈ [0, T ],
Z
R3
ϕ2(t, x)fn(x, t) dx = In(t) n→+∞
−→ 0. (37)
Now, plugging (36) and (37) in (35) we obtain, for all t in [0, T ], Z R3 ϕ(t) (|un(t)|2⋆ 1 |x|)un(t) − (|u(t)| 2⋆ 1 |x|)u(t) ≤ Ckunk2L∞ (0,T ;H1)kun− ukL∞(0,T ;L2(B R))+ kukL∞(0,T ;L2) p In(t) ≤ Ckun− ukL∞(0,T ;L2(B R))+ p In(t) n→+∞ −→ 0. Thus we have proved
(|un|2⋆ 1 |x|)un− (|u| 2⋆ 1 |x|)u n→+∞ −→ 0 in D′((0, T ) × R3).
Therefore, we have all the elements to insure that (un) is converging in a weak
sense towards u which is a solution of (33) in the sense of distributions. We finally have to prove that the limit a of the sequence (an) is a solution of
(34). We already know that (an)n≥0 is bounded in W2,1(0, T ) and that an →
a in L∞(0, T ) and we have in [0,T], m d 2a n dt2 = Z R3−|un(x)| 2 ∇ 1 |x − an| dx+ ∇V1n(an).
On the one hand, omitting again the fixed time t in [0, T ], we have ∇Vn
1(an) − ∇V1(a) = (∇V1n(an) − ∇V1n(a)) + (∇V1n(a) − ∇V1(a))
and of course, since Vn
1 is bounded in H and V1n⇀ V1 weakly in H, we get
Therefore we obtain
∇V1n(an) → ∇V1(a) in D′(0, T ).
On the other hand, using the idea of the proof of Lemma 9 we will prove Z R3|u n(x)|2∇ 1 |x − an| dxn→+∞−→ Z R3|u(x)| 2∇ 1 |x − a|dx in D ′(0, T ).
Actually for all t in [0, T ], we can prove that an integration by parts gives Z R3|u n(t, x)|2∇ 1 |x − an(t)| dx− Z R3|u(t, x)| 2 ∇ 1 |x − a(t)|dx ≤ Z R3|u n|2∇ 1 |x − an| − 1 |x − a| dx + Z R3(|u n|2− |u|2)∇ 1 |x − a|dx ≤ C Z R3 |un||∇un| 1 |x − an|− 1 |x − a| +(|un| + |u|)|u|x − a|2n− u| dx. (38) Since un is bounded in L∞(0, T ; H2), using Cauchy-Schwarz and Hardy’s
in-equality we are able to deal with the first right hand side term. Indeed, Z R3|u n(t)||∇un(t)| 1 |x − an(t)|− 1 |x − a(t)| ≤ Z R3 |un||∇un| |x − an||x − a||an− a| ≤ kunkL2(0,T ;H1)k∇unkL2(0,T ;H1)|an− a|L∞(0,T ) ≤ kunk2L∞ (0,T ;H2 )|an− a|L∞(0,T ) ≤ C |an− a|L∞(0,T ) n→+∞ −→ 0. (39)
Now, since a is bounded on (0, T ), un− u n→+∞
−→ 0 in C([0, T ]; H1
loc), u belongs
to L∞(0, T ; H1) and u
n is bounded in L∞(0, T ; H1) independently of n, then
we obtain in an analogous way as in the proof of Lemma 9 that for all t in [0, T ], Z
R3
(|un(t)| + |u(t)|)|un(t) − u(t)|
|x − a(t)|2 dx= Jn(t) n→+∞
−→ 0. (40) In fact, omitting the time t, we have from Hardy’s inequality
and we can prove (see Lemma 9) that for all t in [0, T ] and for all ε > 0, there exists n0∈ N such that for all n > n0, Jn(t) ≤ ε.
Thus, using (39) and (40) together with (38), we get, for all t in [0, T ], Z R3|u n(t, x)|2∇ 1 |x − an(t)| dx− Z R3|u(t, x)| 2∇ 1 |x − a(t)|dx ≤ C |an− a|L∞(0,T )+ Jn(t) n→+∞ −→ 0 and we finally obtain the awaited result.
We then have proved that a is a solution of (34). As a consequence, the limit (u, a) of (un, an) is a solution in the sense of distribution of system (1).
More-over, since (u, a) belongs to the class W1,∞(0, T ; L2) ∩ L∞(0, T ; H2∩ H 2)×
W2,1(0, T ), then it satisfies the estimate (4) of Theorem 2 and is in fact a strong
solution of system (1).
Hence the proof of Theorem 1.
Remark : First order optimality condition. As we did in the case when the position a of the nucleus is known (Section 2.2.2), we would like to give an optimality condition for the optimal control V1. A first step is to study the
differentiability of the functional
Φ : eH −→ L∞(0, T ; L2(R3)) × L∞(0, T ) V1 7−→ (u(V1), a(V1))),
where eH is an appropriate Hilbert space. One can notice that the lack of proof for the uniqueness of the solution (u, a) of system (1) makes the study of optimality conditions completely formal. It is a first and main obstacle to prove the differentiability with respect to V1 of Φ and of the cost functional
J: (V1, u) 7→ 1 2ku(T ) − u1k 2 L2+ r 2kV1k 2 e H.
Nevertheless, one can obtain a formal derivative of J and an optimal system. We present these following formal results in order to make completely explicit the difficulty encountered in trying to show the differentiability of the mapping Φ : control → state and to give the possibility to make direct computations on approximations of the optimality system after regularization of the singularities. Thus, assuming that we have uniqueness of the solution of system (1) and assuming that Φ is differentiable, if we set DΦ(δV1) = (z, b), then one can prove
that (z(t, x), b(t)) has to satisfy the following coupled system set in R3× (0, T )
i∂tz+ ∆z + V0z+ V1z= − ∂V0 ∂a · b u − δV1u+ (|u| 2⋆ 1 |x|)z + 2(Re(uz) ⋆ 1 |x|)u z(0) = 0, b(0) = 0, db dt(0) = 0, m d 2b dt2 = − Z R3|u| 2 ∇∂V0 ∂a · b − 2 Z R3
where V0 = 1
|x − a|. Thereafter, if J is differentiable with respect to V1, we obtain that the condition DJ(V1, u)[δV1] = 0, ∀δV1∈ eH now reads
Re Z
R3(u(T, x) − u
1(x)) z(T, x) dx + rhV1, δV1iHe= 0. (41)
The main difficulty we encounter when trying to give a meaning to the sys-tem of equations satisfied by the couple (z, b) is of same nature than the one we had when we studied the equations solved by the difference of two solutions of system (1). Indeed, as for the proof of uniqueness which misses in Theorem 2, even in a formal study of the solutions, we have to deal with singularities of type u
|x|2 that we cannot bound with Hardy’s inequality. Moreover, the use of Marcinkiewicz (or Lorentz) spaces as in reference [5] is not directly appropriate here because of the properties of V1.
At last, the following formal adjoint system i∂tp+ ∆p + V0p+ V1p= (|u|2⋆ 1 |x|)p + 2i (Im(up) ⋆ 1 |x|)u− 2iu ∇V0· ̺, p(T ) = u(T ) − u1, m d 2̺ dt2 = − Z R3 ∂V0 ∂a Im(up)− 2 Z R3 Re(u∇u)∂V0 ∂a · ̺ − ∇(∇V1)(a) · ̺, ̺(T ) = 0, d̺ dt(T ) = 0. is such that we have
Re Z R3 z(T ) (u(T ) − u1) = −Im Z T 0 Z R3 δV1up− Z T 0 ̺· ∇δV1(a). (42)
Eventually, if δa denotes the Dirac mass at point a ∈ R3, and using (41) and
(42), we prove that the bilinear optimal control V1 is the solution of a partial
differential equation defined by variational formulation: ∀δV1∈ eH
rhV1, δV1iHe= Z T 0 Z R3 Im(u(t, x)p(t, x)) δV1(t, x) dxdt− Z T 0 ̺(t) · ∇δa(t), δV1(t) dt.
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